How do you find a formula for the sum n terms #sum_(i=1)^n (1+i/n)(2/n)# and then find the limit as #n-oo#?

Question

How do you find a formula for the sum n terms #sum_(i=1)^n (1+i/n)(2/n)# and then find the limit as #n->oo#?

Samantha Adkison · Accepted Answer

# sum_(i=1)^n (1+i/n)(2/n) = (3n+1)/n #
# lim_(n rarr oo)sum_(i=1)^n (1+i/n)(2/n) = 3 #

Let # S_n = sum_(i=1)^n (1+i/n)(2/n) # # :. S_n = sum_(i=1)^n (2/n+(2i)/n^2) # # :. S_n = 2/n sum_(i=1)^n (1) + 2/n^2 sum_(i=1)^n (i)# And using the standard results: # sum_(r=1)^n r = 1/2n(n+1) # We have; # S_n = 2/n(n)+2/n^2*1/2n(n+1) # # :. S_n = 2+(n+1)/n # # :. S_n = ((2n)+(n+1))/n # # :. S_n = (3n+1)/n # Now we examine the behaviour of # S_n # as # n rarr oo #. We have; # S_n = (3n+1)/n # # :. S_n = 3+1/n # # :. lim_(n rarr oo)S_n = lim_(n rarr oo) { 3+1/n } # # :. lim_(n rarr oo)S_n = lim_(n rarr oo) (3) - lim_(n rarr oo)(1/n) # And as #1/n rarr 0# as #n rarr oo#, we have; # :. lim_(n rarr oo)S_n = 3 #

Ezekiel Alexander · Answer

undefined To find the formula for the sum of the given expression, we first need to rewrite it as a Riemann sum. By doing so, we can recognize it as the sum of a sequence and then find its limit as n approaches infinity.

The given expression can be rewritten as:

$$
\frac{2}{n} \sum_{i=1}^{n} \left(1 + \frac{i}{n}\right)
$$

Expanding the sum and applying the formula for the sum of the first n natural numbers, we get:

$$
\frac{2}{n} \left( n + \frac{n(n+1)}{2n} \right)
$$

Simplify the expression:

$$
\frac{2}{n} \left( n + \frac{n+1}{2} \right)
$$

$$
2 + \frac{n+1}{n}
$$

Now, as n approaches infinity, the term $\frac{n+1}{n}$ approaches 1. Therefore, the limit of the expression as n approaches infinity is:

$$
\lim_{n \to \infty} \left(2 + \frac{n+1}{n}\right) = \boxed{2}
$$

How do you find a formula for the sum n terms #sum_(i=1)^n (1+i/n)(2/n)# and then find the limit as #n-&gt;oo#?

How do you find a formula for the sum n terms #sum_(i=1)^n (1+i/n)(2/n)# and then find the limit as #n->oo#?